What is the balanced equation for the combustion of 1.00 mol of camphor, C10H16O(s). I searched the web, but cant find anything, can someone pleas help?

Goodness. Just write the combustion equation and balance.

Balance this

C10H16O+ O2 >>>CO2 + H2O

C10H16O + 10 O2 --> 10 CO2 + 8 H2O

The balanced equation for the combustion of camphor, C10H16O(s), can be determined by following a step-by-step process. Here's how you can balance the equation:

Step 1: Write the skeleton equation for the combustion of camphor:
C10H16O + O2 → CO2 + H2O

Step 2: Balance the carbon atoms by adjusting the coefficient in front of CO2:
C10H16O + O2 → 10CO2 + H2O

Step 3: Balance the hydrogen atoms by adjusting the coefficient in front of H2O:
C10H16O + O2 → 10CO2 + 8H2O

Step 4: Balance the oxygen atoms. Count the total number of oxygen atoms on each side:
On the left side: 1 oxygen atom (from O2)
On the right side: 10 carbon dioxide molecules (each containing 2 oxygen atoms) = 20 oxygen atoms
Plus, 8 water molecules = 16 oxygen atoms

To equalize the number of oxygen atoms on both sides, adjust the coefficient in front of O2 to 19:
C10H16O + 19O2 → 10CO2 + 8H2O

Now the equation is balanced. The balanced equation for the combustion of 1.00 mol of camphor, C10H16O(s), is:
C10H16O + 19O2 → 10CO2 + 8H2O

To find the balanced equation for the combustion of camphor, we need to understand the chemical formula of camphor and the reaction products during combustion.

Camphor has the chemical formula C10H16O, indicating that it contains 10 carbon atoms, 16 hydrogen atoms, and 1 oxygen atom. During combustion, camphor reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

To balance the equation, we need to ensure that there is an equal number of each type of atom on both sides of the equation. Here's how you can balance it:

1. Start by writing the unbalanced equation:
C10H16O + O2 → CO2 + H2O

2. Balance the carbon atoms by placing a coefficient in front of CO2:
C10H16O + O2 → 10CO2 + H2O

3. Balance the hydrogen atoms by placing a coefficient in front of H2O:
C10H16O + O2 → 10CO2 + 8H2O

4. Balance the oxygen atoms by adjusting the coefficient in front of O2:
C10H16O + 21O2 → 10CO2 + 8H2O

So, the balanced equation for the combustion of 1.00 mol of camphor (C10H16O) is:
C10H16O + 21O2 → 10CO2 + 8H2O

Remember, to do this yourself in the future, you first need to know the chemical formula of the substance and understand the reaction that occurs during combustion. Then, you balance the equation by ensuring the number of atoms is the same on both sides.